order that destructors get called?

Stephen Hansen apt.shansen at gmail.invalid
Thu Apr 8 00:16:13 CEST 2010


On 2010-04-07 15:08:14 -0700, Brendan Miller said:
> When doing this, I noticed some odd behaviour. I had code like this:
> 
> def delete_my_resource(res):
>     # deletes res
> 
> class MyClass(object):
>     def __del__(self):
>         delete_my_resource(self.res)
> 
> o = MyClass()
> 
> What happens is that as the program shuts down, delete_my_resource is released
> *before* o is released. So when __del__ get called, delete_my_resource is now
> None.

The first thing Python does when shutting down is go and set the 
module-level value of any names to None; this may or may not cause 
those objects which were previously named such to be destroyed, 
depending on if it drops their reference count to 0.

So if you need to call something in __del__, be sure to save its 
reference for later, so that when __del__ gets called, you can be sure 
the things you need are still alive. Perhaps on MyClass, in its 
__init__, or some such.

> What I'm wondering is if there's any documented order that reference counts
> get decremented when a module is released or when a program terminates.
> 
> What I would expect is "reverse order of definition" but obviously that's not
> the case.

AFAIR, every top level name gets set to None first; this causes many 
things to get recycled. There's no order beyond that, though. 
Namespaces are dictionaries, and dictionaries are unordered. So you 
can't really infer any sort of order to the destruction: if you need 
something to be alive when a certain __del__ is called, you have to 
keep a reference to it.

-- 
--S

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